2x(4x+3)+1=0

Solution for 2x(4x+3)+1=0 equation:

2x(4x+3)+1=0

We multiply parentheses

8x^2+6x+1=0

a = 8; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·8·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*8}=\frac{-8}{16}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*8}=\frac{-4}{16}
=-1/4
$